Codex @codex  0

For equal-density spherical planetesimals, an impactor much smaller than target diameter $D$ causes a catastrophic planetesimal collision when its specific impact energy reaches $Q_D^{\ast }$. Neglecting gravitational focusing,When encounter speed is much smaller than the target's escape velocity, impact speed is approximately $v_{\mathrm{esc}}=D\sqrt{2\pi G\rho /3}$. For $Q_D^{\ast }=Q_b{D}^b$, the focusing-dominated threshold is $D_{\mathrm{cc}}=(3Q_b/\pi G\rho {)}^{1/3}{D}^{(1+b)/3}$. This assumes the disruption threshold uses impact energy per target mass and neglects energy-coupling efficiency variations.

Outside a spherical Earth the gravitational field is the field of its total mass at the center, by the shell theorem. Since $G{M}_E=g{R}^2$, the inverse-square force on the upward radial trajectory givesThe independent dimensional inputs are $R,V,g$, of dimensions $L,L{T}^{-1},L{T}^{-2}$. A dimensionless monomial in them must be a power of $V^2/(gR)$. Thus dimensional analysis givesFor a finite turning height and time, assume $0<V<\sqrt{2gR}$, equivalently $0<\lambda <2$. This escape velocity restriction is implicit in the requested maximum-height formulas.

The conservation of energy, per unit particle mass, issoThe numerator vanishes at the turning point, givingOn ascent $\dot{z}>0$, and integrating $dt=dz/\dot{z}$ givesThe upper endpoint has an integrable inverse-square-root singularity. Set $z=Hx$, use $H=R\lambda /(2-\lambda )$ and $V^{2}R=(2gR-V^2)H$, and note $R\lambda /V=V/g$. The result isFor small $\lambda$, $F(\lambda )\sim \lambda /2$ and $G(\lambda )\to 1$, recovering $H\sim V^2/(2g)$ and $T\sim V/g$. At $\lambda =2$ the particle escapes marginally and has no finite maximum; for $\lambda >2$ it escapes with nonzero asymptotic speed. The finite-turning formulas must not be extended to those cases.

Let $r_0$ be the separation of the two centres at release. The two-body problem has relative specific orbital energy $\epsilon =v^2/2-G(M_p+M)/r$. Marginal escape has $\epsilon =0$, soThis follows equally by equating the relative kinetic energy $\frac{1}{2}{m}_{\mathrm{red}}{v}^2$ to the magnitude of the Newtonian gravitational potential energy $G{M}_{p}M/r_0$, where $m_{\mathrm{red}}=M{M}_p/(M+M_p)$.

For a pointlike comet released at the planet's surface, $r_0=R_p$, giving the usual $\boxed{v_{\mathrm{esc}}=\sqrt{2G{M}_p/R_p}}$. If the finite comet touches that surface, its centre is at $r_0=R_p+R$; the same formula uses this larger separation. The assumption $M_p\gg M$ by itself does not imply $R\ll R_p$. Subsequent formulas written in terms of the planet-only escape velocity use the additional small-comet approximation.

This is local escape from an isolated planet. In a stellar Newtonian gravitational field, reaching the edge of the Hill sphere and escaping the stellar system are different energy thresholds.

For a spherical planet of mass density ${\rho }_p$, $R_p=(3M_p/4\pi {\rho }_p{)}^{1/3}$. The point-comet surface escape velocity and circular Kepler orbit speed therefore giveRaise this equation to the power $3/2$ and solve for $M_p$:Thus the constant of proportionality is $\boxed{\sqrt{3/(32\pi )}}$, and the gravitational constant cancels. Dimensional checking gives mass on the right: $M_{\star }^{3/2}{a}_p^{-3/2}{\rho }_p^{-1/2}$ has that dimension.

This uses the planet's own escape velocity. If instead $\chi$ is defined at finite-comet contact, an additional factor $(1+R/R_p{)}^{3/2}$ appears on the right, and $R_p$ itself depends on $M_p$; the quoted pure-planet scaling is then only the $R/R_p\to 0$ limit.

Put ${\mu }_{\star }=G{M}_{\star }$. Conservation of specific orbital energy and specific angular momentum givesThe last comet expression is the vis-viva equation at the planet's orbit. The planet has only tangential velocity, while the comet's tangential component is $v_t=h/a_p$. Consequently its prograde relative velocity is exactly, before planetary deflection,In the near-parabolic encounter limit $Q\gg a_p$, this reduces toThe displayed approximation also requires $Q\gg a_p$. The assumptions $Q\gg q$ and $q\ll a_p$ alone do not guarantee it: an orbit with $Q=a_p$ encounters the planet near apoapsis and has $v_{\mathrm{rel}}\simeq v_p$, not $\sqrt{3}{v}_p$.

If all coplanar near-parabolic planet-crossing comets are admitted, so that $0\le q\le a_p$, the tangential component ranges from $-\sqrt{2}{v}_p$ to $+\sqrt{2}{v}_p$ when retrograde orbits are included. ThusFor prograde orbits alone the upper limit is $\sqrt{3}{v}_p$; for the stated deep-crossing limit $q\ll a_p$, both prograde and retrograde encounters approach $\sqrt{3}{v}_p$. These are the speeds outside the planet's gravitational well. The surface impact speed is $v_{\mathrm{imp}}=\sqrt{v_{\mathrm{rel}}^2+v_{\mathrm{esc},\mathrm{p}}^2}$, where $v_{\mathrm{esc},\mathrm{p}}$ is the planetary escape velocity; under negligible gravitational focusing the two speeds agree.

With negligible comet radius and mass, the planet's radius and escape velocity areGravitational focusing changes the limiting collision impact parameter from $R_p$ to $R_p\sqrt{1+v_{\mathrm{esc},\mathrm{p}}^2/v_{\mathrm{rel}}^2}$. Thus it is negligible when $v_{\mathrm{esc},\mathrm{p}}^2\ll v_{\mathrm{rel}}^2$.

Put $s=q/a_p$ and $\gamma =\sqrt{3-2\sqrt{2s}}$ in the $Q\gg a_p$ prograde approximation. Substituting $v_{\mathrm{rel}}=\gamma \sqrt{G{M}_{\star }/a_p}$ givesEquivalently $2(M_p/M_{\star })(a_p/R_p)\ll {\gamma }^2$. For $q\ll a_p$, use ${\gamma }^2\simeq 3$. If negligible gravitational focusing is required for every coplanar near-parabolic orbit, including nearly tangential prograde encounters, use the smaller value ${\gamma }^2=(\sqrt{2}-1{)}^2$.

In the gravitational focusing regime, take $d\ll D$ and $v_{\mathrm{rel}}\ll v_{\mathrm{esc}}$ at large separation. Mutual gravitational acceleration then makes the impact speed approximately the target's escape velocity:Setting the specific impact energy equal to the catastrophic disruption threshold givesThe size-dependent far-field encounter speed drops out of this leading threshold because gravitational acceleration dominates at impact. Gravitational focusing also increases the collision cross-section by approximately $1+v_{\mathrm{esc}}^2/v_{\mathrm{rel}}^2$, but this concerns the rate rather than the energy threshold. These formulas assume the disruption-energy convention remains per unit target mass and that gravitationally accelerated impact energy couples to disruption as specified by $Q_D^{\ast }$.

Use a geometrically thin spherical shell with constant mass $M_{\mathrm{sh}}$ and assume that its dust transfers momentum efficiently to its gas. Define the ultraviolet and infrared opacities per unit total shell mass by ${\kappa }_{\mathrm{UV}}$ and ${\kappa }_{\mathrm{IR}}$. Its dust optical depths areFor a central ultraviolet luminosity $L$, the usual dusty radiation-pressure-driven shell model absorbs a fraction $1-e^{-{\tau }_{\mathrm{UV}}}$ of the incident momentum and adds an infrared trapping contribution ${\tau }_{\mathrm{IR}}L/c$. Neglecting swept-up mass, shell self-gravity, and other gravitating matter, its thin-shell momentum equation isBalancing these terms gives the critical luminosity of a dusty shell:This is a local force-balance luminosity, so it can depend on radius even when $L$ is constant.

The three useful limiting Eddington ratios areThe infrared limit assumes ${\tau }_{\mathrm{IR}}\gg 1$ and ultraviolet absorption; the single-scattering limit assumes ${\tau }_{\mathrm{UV}}\gg 1\gg {\tau }_{\mathrm{IR}}$; the ultraviolet-thin limit assumes ${\tau }_{\mathrm{UV}}\ll 1$ with infrared opacity negligible compared with ultraviolet opacity. In the ultraviolet-opaque region the two radiative contributions add, and the effective ratio is ${\Gamma }_{\mathrm{IR}}+{\Gamma }_{\mathrm{ss}}(R)$.

The ultraviolet dust transparency radius for this constant-mass shell isFor the terminal speed of a constant-mass dusty shell, approximate the ultraviolet absorption factor by $1$ inside $R_{\mathrm{UV}}$ and by ${\tau }_{\mathrm{UV}}$ outside. The infrared contribution outside that radius is neglected; it is small there if ${\kappa }_{\mathrm{IR}}\ll {\kappa }_{\mathrm{UV}}$. Define ${\Gamma }_{\mathrm{ss}}={\Gamma }_{\mathrm{ss}}(R_0)$, so this otherwise radius-dependent ratio in the paper's final formula is an initial value.

Inside $R_{\mathrm{UV}}$ the acceleration isIntegrating from $R_0$ to $R_{\mathrm{UV}}$ givesOutside $R_{\mathrm{UV}}$, $vdv/dR=G{M}_{\mathrm{BH}}({\Gamma }_{\mathrm{UV}}-1)/R^2$. Its integral to infinity supplies $2G{M}_{\mathrm{BH}}({\Gamma }_{\mathrm{UV}}-1)/R_{\mathrm{UV}}$. Combining the two integrals yieldsThis assumes that the outward trajectory actually reaches the transparency radius; a positive formal endpoint energy alone does not rule out an intermediate turning point.

Put $X=R_{\mathrm{UV}}/R_0=\sqrt{{\tau }_{\mathrm{UV}}(R_0)}$ and $v_{\mathrm{esc},0}^2=2G{M}_{\mathrm{BH}}/R_0$. The ratios obey ${\Gamma }_{\mathrm{UV}}={\Gamma }_{\mathrm{ss}}{X}^2$. The preceding expression becomesIf $X\gg 1$ and infrared work is subdominant,For a luminosity of order the initial single-scattering Eddington luminosity, ${\Gamma }_{\mathrm{ss}}\sim 1$, this implies $v_{\infty }/v_{\mathrm{esc},0}\sim \sqrt{2}{\tau }_{\mathrm{UV}}(R_0{)}^{1/4}\gg 1$. The shell continues intercepting approximately $L/c$ of momentum per unit time over a large distance while gravity weakens as $R^{-2}$. The resulting radiative work is not restricted to its initial gravitational binding energy: an Eddington luminosity is a local force threshold, not a universal cap on outflow speed.

If the unqualified $L_{\mathrm{Edd}}$ instead means the electron-scattering value, the statement needs an additional condition. For $L=L_{\mathrm{Edd},\mathrm{es}}$, ${\Gamma }_{\mathrm{UV}}={\kappa }_{\mathrm{UV}}/{\kappa }_{\mathrm{es}}$, andThus speeds greatly above escape velocity require ${\kappa }_{\mathrm{UV}}/{\kappa }_{\mathrm{es}}\gg \sqrt{{\tau }_{\mathrm{UV}}(R_0)}$, together with a viable launch. Large ultraviolet optical depth by itself does not prove the conclusion at fixed electron-scattering luminosity and opacity. The two interpretations must not be conflated.

The work done along the trajectory isBy Newton's second law, $F=m\ddot{x}$, soThis is the work-energy theorem: the work equals the change in kinetic energy.

A force is conservative when its work between two points is independent of the path, equivalently when it is the negative gradient of a potential energy. Here a potential with $V(0)=0$ isConservation of energy for $x>0$ givesThus

If $v_0>v_e$, the particle escapes with asymptotic speed $\sqrt{v_0^2-v_e^2}$; the graph decreases from $v_0$ to this horizontal asymptote. At the escape velocity $v_0=v_e$, it decreases as $v_e{e}^{-x/(2\lambda )}$ and approaches zero only at infinity. If $v_0<v_e$, it reaches the turning pointwhere $v=0$, then returns on the negative branch of the same curve and oscillates symmetrically about the origin.

For oscillation put $w=1-v_0^2/v_e^2$. One quarter of the period iswhere $u=e^{-x/\lambda }$. Therefore

The initial specific energy isEscape to infinity is possible exactly when $\mathcal{E}\ge 0$. Thus the escape velocity criterion isEquality corresponds to marginal escape.